Sum of minimum number of vertices

We are considering the sum over the set $2^{\lbrace 0,1 \rbrace} = \lbrace \varnothing, \lbrace 0 \rbrace, \lbrace 1 \rbrace, \lbrace 0,1 \rbrace \rbrace$ . This just means we are summing the minima over the intervals $(0,1), [0,1), (0,1],$ and $[0,1]$ .

In the interval $(0,1)$ , there is a point in time $t_{r}$ where the cube is rectified, becoming a cuboctahedron. At this point, $V(t_{r}) = 12$ , as compared to $V(t) = 24$ elsewhere. As such,

$\min_{t \in (0,1)} V(t) = 12$

At $t=0$ , we have a cube with $V(0) = 8$ . At $t=1$ , we have an octahedron with $V(1) = 6$ . Thus,

$\min_{t \in [0,1)} V(t) = 8$

$\min_{t \in (0,1]} V(t) = 6$

$\min_{t \in [0,1]} V(t) = 6$

Hence,

$\sum_{S \in 2^{\lbrace 0,1 \rbrace}} \min_{t \in (0,1) \cup S} V(t) = 12+8+6+6 = \boxed{32}$